Given: vertex $A(2,-3).$ and directrix $x-4y-48=0$.............$(i)$

We know that Axis of parabola is $\perp$ to directrix and

it passes through focus and vertex of the parabola.

Equation of a line $\perp$ to the line $ax+by+c=0$ is given by $bx-ay+c_1=0$.

$\therefore$ The equation of the axis is $-4x-y+c=0$

or $ 4x+y+c=0$

But vertex $ (2,-3)$ lies on the axis of the parabola.

$\therefore\:$It satisfies the equation of the axis. $\Rightarrow\:8-3+c=0$ or $ c=-5$

$\therefore$ The equation of axis is $4x+y-5=0$..............$(ii)$

Foot of $\perp$ from vertex to directrix(A') is got by solving the equations of axis and directrix.

By solving $(i)\:\:and\:\:(ii)$ we get the foot of $\perp$ of vertex $(A')$ on directrix which is

$(x-4y-48)+4(4x+y-5)=0$

$\Rightarrow\:17x-68=0$ or $x=4\:\:and\:\:y=-11$

$\therefore A'(4,-11)$

We know that Vertex(A) is the mid point of focus(S) and foot of $\perp$ $(A')$

Let focus be $S(x,y)$

$\therefore \:\big(\large\frac{x+4}{2},\frac{y-11}{2}\big)=(2,-3)$

$\Rightarrow\:x=0\:\:and\:\:y=5$

$i.e., focus\: (S)=(0,5)$

We know that equation of the parabola is equation of locus of a point which moves

such that its distance from focus = its distance from directrix.

$\therefore$ Equation of the parabola is:

Let $P(x,y)$ be any point on the parabola.

As per the def. of parabola Distance PS=Distance P$\rightarrow\:Directrix (i)$

$\Rightarrow\:\sqrt{(x-0)^2+(y-5)^2}=\large\frac{x-4y-48}{\sqrt {1+16}}$

Squaring on both the sides we get

$17x^2+17y^2+425-170y=x^2+16y^2+48^2-8xy+384y-96x$

$\Rightarrow\: 16x^2+y^2+8xy+96x-554y-1879=0$